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Choose the Best Answer · Q17

Q.Br Br∣∣\underset{|\qquad|}{\text{Br}\quad\ \text{Br}}
CH2-CH2 →(A)\xrightarrow{(A)} CH≡\equivCH (with Br on each CH2 carbon), where A is,

(a) Zn
(b) Conc. H2SO4
(c) alc. KOH
(d) dil. H2SO4
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Step 1. The starting material, BrCH2-CH2Br (1,2-dibromoethane), must lose TWO molecules of HBr to reach the fully unsaturated product CH≡\equivCH (ethyne) -- two successive dehydrohalogenation (elimination) steps.

Step 2. Elimination of HX from a haloalkane is achieved with a base, and alcoholic KOH is the standard reagent used throughout this chapter for exactly this kind of dehydrohalogenation (e.g. converting a monohalide to an alkene, or -- with enough base/heating -- carrying a vicinal dihalide onward to an alkyne). …

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